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Unival Trees in Go
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package main | |
//Given a tree, determine how many of its subtrees are inival trees. | |
//A unival tree is a tree where all nodes have the same value. | |
//An empty tree is a single unival tree. | |
//A tree with one node is a single unival tree. | |
//Assumptions | |
// - A tree whose value equals one of its children and the other child is nil is not a unival tree. | |
// - nil children do not contribute to the count of unival trees. | |
import "fmt" | |
type univalNode struct { | |
value int | |
left *univalNode | |
right *univalNode | |
} | |
func countUnivalSubTrees(root *univalNode) (isUnival bool, count int) { | |
isUnival, count = true, 1 | |
if root == nil { | |
return | |
} else if root.left == nil && root.right == nil { | |
return | |
} | |
leftIsUnival, leftCount := false, 0 | |
if root.left != nil { | |
leftIsUnival, leftCount = countUnivalSubTrees(root.left) | |
} | |
rightIsUnival, rightCount := false, 0 | |
if root.right != nil { | |
rightIsUnival, rightCount = countUnivalSubTrees(root.right) | |
} | |
isUnival = leftIsUnival && rightIsUnival && root.value == root.left.value && root.value == root.right.value | |
count = leftCount + rightCount | |
if isUnival { | |
count++ | |
} | |
return | |
} | |
func main() { | |
fmt.Println(countUnivalSubTrees(nil)) | |
fmt.Println(countUnivalSubTrees(&univalNode{ | |
value: 0, | |
left: nil, | |
right: nil, | |
})) | |
fmt.Println(countUnivalSubTrees(&univalNode{ | |
value: 0, | |
left: &univalNode{ | |
value: 0, | |
left: nil, | |
right: nil, | |
}, | |
right: &univalNode{ | |
value: 1, | |
left: nil, | |
right: nil, | |
}, | |
})) | |
fmt.Println(countUnivalSubTrees(&univalNode{ | |
value: 0, | |
left: &univalNode{ | |
value: 0, | |
left: &univalNode{ | |
value: 0, | |
left: &univalNode{ | |
value: 0, | |
left: &univalNode{ | |
value: 0, | |
left: nil, | |
right: nil, | |
}, | |
right: nil, | |
}, | |
right: nil, | |
}, | |
right: nil, | |
}, | |
right: nil, | |
})) | |
} |
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